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Linear Algebra : Linear Algebra

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Example Questions

Example Question #13 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-20&-7\\-14&12\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-22.81;\lambda_{2}=14.81$

$\lambda_{1}=-26.81;\lambda_{2}=10.81$

$\lambda_{1}=-29.81;\lambda_{2}=7.81$

$\lambda_{1}=-23.81;\lambda_{2}=13.81$

Correct answer:

$\lambda_{1}=-22.81;\lambda_{2}=14.81$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-20&-7\\-14&12\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 20&-7\\-14&12 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{From that, we can find the eigenvalues:}\\&8\lambda + \lambda ^{2} - 338\\&\lambda_{1}=-22.81;\lambda_{2}=14.81\end{align*}$

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Example Question #14 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-2&-17&-11\\17&-14&13\\2&20&-17\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=1.17+17.26i;\lambda_{2}=1.17-17.26i;\lambda_{3}=-29.33$

$\lambda_{1}=-0.83+17.26i;\lambda_{2}=-0.83-17.26i;\lambda_{3}=-31.33$

$\lambda_{1}=5.17+17.26i;\lambda_{2}=5.17-17.26i;\lambda_{3}=-25.33$

$\lambda_{1}=-9.83+17.26i;\lambda_{2}=-9.83-17.26i;\lambda_{3}=-40.33$

Correct answer:

$\lambda_{1}=-0.83+17.26i;\lambda_{2}=-0.83-17.26i;\lambda_{3}=-31.33$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-2&-17&-11\\17&-14&13\\2&20&-17\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 2&-17&-11\\17&- \lambda - 14&13\\2&20&- \lambda - 17\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{From that, we can find the eigenvalues:}\\&(-1)\cdot (351\lambda + 33\lambda ^{2} + \lambda ^{3} + 9359)\\&\lambda_{1}=-0.83+17.26i;\lambda_{2}=-0.83-17.26i;\lambda_{3}=-31.33\end{align*}$

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Example Question #15 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-4&-10\\12&-3\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-10.50+10.94i;\lambda_{2}=-10.50-10.94i$

$\lambda_{1}=0.50+10.94i;\lambda_{2}=0.50-10.94i$

$\lambda_{1}=-3.50+10.94i;\lambda_{2}=-3.50-10.94i$

$\lambda_{1}=-0.50+10.94i;\lambda_{2}=-0.50-10.94i$

Correct answer:

$\lambda_{1}=-3.50+10.94i;\lambda_{2}=-3.50-10.94i$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-4&-10\\12&-3\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 4&-10\\12&- \lambda - 3\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&7\lambda + \lambda ^{2} + 132\\&\lambda_{1}=-3.50+10.94i;\lambda_{2}=-3.50-10.94i\end{align*}$

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Example Question #16 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-15&14&5\\-6&1&-4\\-17&-11&-15\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-15.74+12.27i;\lambda_{2}=-15.74-12.27i;\lambda_{3}=2.49$

$\lambda_{1}=-24.74+12.27i;\lambda_{2}=-24.74-12.27i;\lambda_{3}=-6.51$

$\lambda_{1}=-14.74+12.27i;\lambda_{2}=-14.74-12.27i;\lambda_{3}=3.49$

$\lambda_{1}=-10.74+12.27i;\lambda_{2}=-10.74-12.27i;\lambda_{3}=7.49$

Correct answer:

$\lambda_{1}=-15.74+12.27i;\lambda_{2}=-15.74-12.27i;\lambda_{3}=2.49$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-15&14&5\\-6&1&-4\\-17&-11&-15\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 15&14&5\\-6&1 - \lambda&-4\\-17&-11&- \lambda - 15\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Knowing this, we can expand and solve:}\\&(-1)\cdot (320\lambda + 29\lambda ^{2} + \lambda ^{3} - 992)\\&\lambda_{1}=-15.74+12.27i;\lambda_{2}=-15.74-12.27i;\lambda_{3}=2.49\end{align*}$

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Example Question #17 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-7&16\\-5&-16\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-8.50+7.73i;\lambda_{2}=-8.50-7.73i$

$\lambda_{1}=-11.50+7.73i;\lambda_{2}=-11.50-7.73i$

$\lambda_{1}=-7.50+7.73i;\lambda_{2}=-7.50-7.73i$

$\lambda_{1}=-4.50+7.73i;\lambda_{2}=-4.50-7.73i$

Correct answer:

$\lambda_{1}=-11.50+7.73i;\lambda_{2}=-11.50-7.73i$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-7&16\\-5&-16\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 7&16\\-5&- \lambda - 16\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{From that, we can find the eigenvalues:}\\&23\lambda + \lambda ^{2} + 192\\&\lambda_{1}=-11.50+7.73i;\lambda_{2}=-11.50-7.73i\end{align*}$

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Example Question #11 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Find the eigenvalues for the matrix }\\&A=\begin{bmatrix}3&-18&-11\\-6&13&-20\\-19&-14&6\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-22.30;\lambda_{2}=22.15+5.95i;\lambda_{3}=22.15-5.95i$

$\lambda_{1}=-27.30;\lambda_{2}=17.15+5.95i;\lambda_{3}=17.15-5.95i$

$\lambda_{1}=-25.30;\lambda_{2}=19.15+5.95i;\lambda_{3}=19.15-5.95i$

$\lambda_{1}=-29.30;\lambda_{2}=15.15+5.95i;\lambda_{3}=15.15-5.95i$

Correct answer:

$\lambda_{1}=-22.30;\lambda_{2}=22.15+5.95i;\lambda_{3}=22.15-5.95i$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}3&-18&-11\\-6&13&-20\\-19&-14&6\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}3 - \lambda&-18&-11\\-6&13 - \lambda&-20\\-19&-14&6 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Using this, we can solve for our eigenvalues:}\\&(-1)\cdot (\lambda ^{3} - 22\lambda ^{2} - 462\lambda + 11735)\\&\lambda_{1}=-22.30;\lambda_{2}=22.15+5.95i;\lambda_{3}=22.15-5.95i\end{align*}$

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Example Question #19 : Eigenvalues And Eigenvectors

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-13&-5&5\\11&-17&18\\11&-1&-3\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-16.08+3.47i;\lambda_{2}=-16.08-3.47i;\lambda_{3}=-27.84$

$\lambda_{1}=-7.08+3.47i;\lambda_{2}=-7.08-3.47i;\lambda_{3}=-18.84$

$\lambda_{1}=-12.08+3.47i;\lambda_{2}=-12.08-3.47i;\lambda_{3}=-23.84$

$\lambda_{1}=-5.08+3.47i;\lambda_{2}=-5.08-3.47i;\lambda_{3}=-16.84$

Correct answer:

$\lambda_{1}=-7.08+3.47i;\lambda_{2}=-7.08-3.47i;\lambda_{3}=-18.84$

Explanation:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-13&-5&5\\11&-17&18\\11&-1&-3\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 13&-5&5\\11&- \lambda - 17&18\\11&-1&- \lambda - 3\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Knowing this, we can expand and solve:}\\&(-1)\cdot (329\lambda + 33\lambda ^{2} + \lambda ^{3} + 1172)\\&\lambda_{1}=-7.08+3.47i;\lambda_{2}=-7.08-3.47i;\lambda_{3}=-18.84\end{align*}$

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Example Question #20 : Eigenvalues And Eigenvectors

Give the characteristic polynomial of the matrix

$A = \begin{bmatrix} \frac{1}{2}& \frac{2}{3}\\ \\ -\frac{1}{4}& \frac{1}{3} \end{bmatrix}$

Possible Answers:

$\lambda ^{2} - \frac{5}{6} \lambda$

$\lambda ^{2} + \frac{5}{6} \lambda$

$\lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{3}$

$\lambda ^{2} + \frac{5}{6} \lambda + \frac{1}{6}$

$\lambda ^{2} + \frac{5}{6} \lambda + \frac{1}{3}$

Correct answer:

$\lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{3}$

Explanation:

The characteristic polynomial of a square matrix can be derived as follows:

Determine $\det( \lambda I - A)$ , using the identity matrix with the same dimensions as (two by two):

$\lambda I - A$

$= \lambda \begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix} - \begin{bmatrix} \frac{1}{2}& \frac{2}{3}\\ \\ -\frac{1}{4}& \frac{1}{3} \end{bmatrix}$

$= \begin{bmatrix} \lambda& 0\\ 0& \lambda \end{bmatrix} - \begin{bmatrix} \frac{1}{2}& \frac{2}{3}\\ \\ -\frac{1}{4}& \frac{1}{3} \end{bmatrix}$

Subtract the matrices by subtracting elementwise:

$=\begin{bmatrix} \lambda - \frac{1}{2}& 0- \frac{2}{3}\\ \\ 0 - \left (-\frac{1}{4} \right )& \lambda - \frac{1}{3} \end{bmatrix}$

$=\begin{bmatrix} \lambda - \frac{1}{2}& - \frac{2}{3}\\ \\ \frac{1}{4} & \lambda - \frac{1}{3} \end{bmatrix}$

Find the determinant of this matrix by taking the product of the upper left-to lower right diagonal and subtracting the product of the upper right-to-lower left diagonal:

$\det( \lambda I - A)$

$=\begin{vmatrix} \lambda - \frac{1}{2}& - \frac{2}{3}\\ \\ \frac{1}{4} & \lambda - \frac{1}{3} \end{vmatrix}$

$=\left ( \lambda - \frac{1}{2} \right )\left ( \lambda - \frac{1}{3} \right ) - \left ( - \frac{2}{3}\right ) \left ( \frac{1}{4} \right )$

$= \lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{6} - \left ( - \frac{1}{6}\right )$

$= \lambda ^{2} - \frac{5}{6} \lambda + \frac{1}{3}$ ,

the correct choice.

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Example Question #471 : Linear Algebra

True or false:

$\begin{bmatrix} 1\\ 2 \end{bmatrix}$ is an eigenvector of the matrix $\begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

Possible Answers:

True

False

Correct answer:

True

Explanation:

A vector is an eigenvector of a matrix if and only if there exists a scalar value $\lambda$ - an eigenvalue - such that

$A v = \lambda v$ ;

or, equivalently, A v must be a scalar multiple of .

Letting $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ and $v= \begin{bmatrix} 1\\ 2 \end{bmatrix}$ , find A v by multiplying each row of by - that is, multiplying each element in each row in by the corresponding element in . This is

$Av = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}\begin{bmatrix} 1\\ 2 \end{bmatrix}$

$= \begin{bmatrix} 4(1)+ 1(2)\\ 2(1)+ 5(2) \end{bmatrix}$

$= \begin{bmatrix}4+2\\ 2+10 \end{bmatrix}$

$= \begin{bmatrix}6\\ 12 \end{bmatrix}$

Since $6 = 6 \cdot 1$ and $12 = 6 \cdot 2$ , it follows that

Av = 6v .

Therefore, such a $\lambda$ exists (and is equal to 6), and $v= \begin{bmatrix} 1\\ 2 \end{bmatrix}$ is indeed an eigenvector of $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

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Example Question #21 : Eigenvalues And Eigenvectors

True or false:

$\begin{bmatrix} 1\\ 1 \end{bmatrix}$ is an eigenvector of the matrix $\begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

Possible Answers:

True

False

Correct answer:

False

Explanation:

A vector is an eigenvector of a matrix if and only if there exists a scalar value $\lambda$ - an eigenvalue - such that

$A v = \lambda v$ ;

or, equivalently, A v must be a scalar multiple of .

Letting $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ and $v= \begin{bmatrix} 1\\ 1 \end{bmatrix}$ , find A v by multiplying each row of by - that is, multiplying each element in each row in by the corresponding element in . This is

$Av = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix}$

$= \begin{bmatrix} 4(1)+ 1(1)\\ 2(1)+ 5(1) \end{bmatrix}$

$= \begin{bmatrix}4+1\\ 2+5\end{bmatrix}$

$= \begin{bmatrix}5\\ 7\end{bmatrix}$

$5 = 5 \cdot 1$ , but $7= 7 \cdot 1$ . Therefore, there cannot exist $\lambda$ such that $A v = \lambda v$ . This means that $v= \begin{bmatrix} 1\\ 1 \end{bmatrix}$ is not an eigenvector of $A = \begin{bmatrix} 4& 1\\ 2& 5 \end{bmatrix}$ .

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #13 : Eigenvalues And Eigenvectors

Example Question #14 : Eigenvalues And Eigenvectors

Example Question #15 : Eigenvalues And Eigenvectors

Example Question #16 : Eigenvalues And Eigenvectors

Example Question #17 : Eigenvalues And Eigenvectors

Example Question #11 : Eigenvalues And Eigenvectors

Example Question #19 : Eigenvalues And Eigenvectors

Example Question #20 : Eigenvalues And Eigenvectors

Example Question #471 : Linear Algebra

Example Question #21 : Eigenvalues And Eigenvectors

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